Answer :
Answer:
See explanation below
Explanation:
The question is incomplete, however, in the picture 1 attached you have the whole question. Picture 2 is the answer of the question.
Now, both of the reactions use the same reactants but in different order. You may say, it would be the same product, cause the factor's order does not affect the final product does it?, actually in this case, it does because toluene have a aromatic electrofilic substitution and the methyl group is a weak activating group, so, this take effect in the position which the substitute will go in the aromatic ring.
In the first reaction, we have the KMnO₄ reacting first. When this happens, occurs an oxydation of the methyl group to convert it to the COOH group. This group is deactivating the molecule in meta direction, so, when the second reactant goes in, the bromine, it will go to the meta position of the ring and the final product would be the 1,3 substitued product.
In the second reaction, as it was said before, the methyl group activate the molecule, and it's orto, para director, so, the bromine will react and go to the orto or para position. In this case, the majority product will be with the para position because of steric hindrance of the methyl in the position 1, 2. Then, the oxydation reaction occurs and the COOH enters again, the final product is similar to the product of the first reaction, but it will be the 1,4 product. Picture 2 has the answer.
